Problem: As you very well know, this year's funkiest numbers are so called triangular numbers (that is, integers that are representable as <image>, where k is some positive integer), and the coolest numbers are those that are representable as a sum of two triangular numbers.

A well-known hipster Andrew adores everything funky and cool but unfortunately, he isn't good at maths. Given number n, help him define whether this number can be represented by a sum of two triangular numbers (not necessarily different)!

Input

The first input line contains an integer n (1 ≤ n ≤ 109).

Output

Print "YES" (without the quotes), if n can be represented as a sum of two triangular numbers, otherwise print "NO" (without the quotes).

Examples

Input

256


Output

YES


Input

512


Output

NO

Note

In the first sample number <image>.

In the second sample number 512 can not be represented as a sum of two triangular numbers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
#pragma GCC optimize("unroll-loops")
using namespace std;
const long long MOD = 998244353;
const long long INF = 1e18;
const long long N = 100000;
void solve() {
  long long n;
  cin >> n;
  for (long long i = 1; (i * (i + 1)) < 2 * n; i++) {
    bool ok = 0;
    long long lo = 1, hi = N, m, k, ele;
    ele = n - ((i * (i + 1)) / 2);
    while (lo <= hi) {
      m = (lo + hi) / 2;
      k = ele - ((m * (m + 1)) / 2);
      if (k < 0)
        hi = m - 1;
      else if (k > 0)
        lo = m + 1;
      else {
        ok = 1;
        break;
      }
    }
    if (ok) {
      cout << "YES";
      return;
    }
  }
  cout << "NO";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  for (int it = 1; it <= t; it++) {
    solve();
  }
  return 0;
}